Question

прошу о помощи. алгебра 11 класс. пределы​

Answer 1

lim(x->3) (x^2 - 5x + 6) / (x^3 - 27) = lim(x->3) (x-3)(x-2)/(x-3)(x^2 + 3x + 9) = lim(x->3) (x-2)/(x^2 + 3x + 9) = 1/(9+9+9) = 1/27

Answer 2

$\lim\limits _{x \to 3}\frac{x^2-5x+6}{x^3-27}=\lim\limits _{x \to 3}\frac{(x-3)(x-2)}{(x-3)(x^2+3x+9)}=\lim\limits _{x \to 3}\frac{x-2}{x^2+3x+9}=\frac{3-2}{9+9+9}=\frac{1}{27}\\\\\\\star \; \; x^2-5x+6=0\; \; \to \; \; x_1=3\; ,\; x_2=2\; \; (teorema\; Vieta)\; \; \to \\\\x^2-5x+6=(x-3)(x-2)\; \; \star$